Question: Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that
\[f(f(x - y)) = f(x) f(y) - f(x) + f(y) - xy\]for all $x,$ $y.$  Find the sum of all possible values of $f(1).$
Answer: Let $a = f(0)$ and $b = f(f(0))$.  Setting $y = x$ in the given equation, we get
\[[f(x)]^2 - x^2 = b \quad (1)\]for all $x$.  In particular, for $x = 0$, $a^2 = b$.

Setting $y = 0$ in the given equation, we get
\[f(f(x)) = (a - 1) f(x) + a \quad (2)\]for all $x$.

Substituting $f(x)$ for $x$ in equation (1), we get
\[[f(f(x))]^2 - [f(x)]^2 = b.\]But from equation (2), $[f(f(x))]^2 = [(a - 1) f(x) + a]^2 = (a^2 - 2a + 1) [f(x)]^2 + 2a(a - 1) f(x) + a^2$, so
\[(a^2 - 2a) [f(x)]^2 + 2a(a - 1) f(x) = af(x) [(a - 2) f(x) + 2(a - 1)] = 0\]for all $x$.

If $a \neq 0$, then
\[f(x) [(a - 2) f(x) + 2(a - 1)] = 0\]for all $x$, so $f(x)$ attains at most two different values.  But by equation (1), this cannot be the case.

Hence, $a = 0$, then $b = 0$, so from equation (1),
\[[f(x)]^2 = x^2,\]which means $f(x) = x$ or $f(x) = -x$ for all $x$.

Let $x$ be a value such that $f(x) = x$.  Then $f(f(x)) = f(x) = x$, so by equation (2), $x = -x$, or $x = 0$.  Hence, the only value of $x$ such that $f(x) = x$ is $x = 0$.  Therefore, $f(x) = -x$ for all $x$.  It is easy to check that this solution works.

Therefore, the sum of all possible values of $f(1)$ is $\boxed{-1}.$